Google just updated their google voice platform. Which seems to directly correlate when my googlevoice login stopped working.
I have tried the following:
allowing captcha as suggested here (pygooglevoice-login-error)
Adapting a 2.7 solution here with no luck Python Google Voice
Logging out of my session that is voice.logout()
Uninstalled pygooglevoice and reinstalled.
Tried a different google voice account.
This code was working perfectly up until the google voice website makeover.
python 3.5.2 windows Server2012R2
from googlevoice import Voice
from googlevoice.util import input
voice = Voice()
voice.login(email='email#gmail.com', passwd='mypassword')
def sendText(phoneNumber,text):
try:
voice.send_sms(phoneNumber, text)
except Exception:
pass
sendText(phoneNumber=[aaabbbcccc],text="Hello from Google Voice!")
voice.logout()
Error Log:
Traceback (most recent call last):
File voice.py, line 95, in login
assert self.special
AssertionError
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
line 7, in <module>
voice.login(email='********', passwd='*******')
File voice.py, line 97, in login
raise LoginError
googlevoice.util.LoginError
I've got the same issue. It looks like the page being sent back is a drastically different, javascript/ajax solution than what was sent before.
I've been messing with it a bit and tracked it to the missing the "special" session token that was included before. PyGoogleVoice is searching for the string literal "_rnr_se" within the page HTML sent back from google to scrape the session value. That string is not found, which causes it to think the login failed. From what I can tell, PGV needs that token to make the url/function calls back to imitate the web client.
There's a javascript function that's retrieving that variable now, instead of it being passed back, hardcoded in the HTML page.
gc.net.XhrManager = function(xsrfToken, notification, loadNotification) {
goog.events.EventTarget.call(this);
this.xsrfToken_ = xsrfToken;
this.notification_ = notification;
this.loadNotification_ = loadNotification;
this.logger_ = goog.debug.Logger.getLogger("gc.Xhr");
this.xhrManager_ = new goog.net.XhrManager(0);
this.activeRequests_ = new goog.structs.Map;
this.eventHandler_ = new goog.events.EventHandler(this);
this.eventHandler_.listen(this.xhrManager_, goog.net.EventType.SUCCESS, this.onRequestSuccess_);
this.eventHandler_.listen(this.xhrManager_, goog.net.EventType.ERROR, this.onRequestError_);
};
And then when making calls, it's using the value like so:
gc.net.XhrManager.prototype.sendPost = function(id, url, queryData, opt_successCallback, opt_errorCallback) {
this.sendAnalyticsEvent_(url, queryData);
id = goog.string.buildString(id, this.idGenerator_.getNextUniqueId());
if (goog.isDefAndNotNull(queryData) && !(queryData instanceof goog.Uri.QueryData)) {
throw Error("queryData parameter must be of type goog.Uri.QueryData");
}
var uri = new goog.Uri(url), completeQueryData = queryData || new goog.Uri.QueryData;
completeQueryData.set("_rnr_se", this.xsrfToken_);
this.activeRequests_.set(id, {queryData:completeQueryData, onSuccess:opt_successCallback, onError:opt_errorCallback});
this.xhrManager_.send(id, uri.toString(), "POST", completeQueryData.toString());
};
I figured I'd share my findings so others can help tinker with the new code and figure out how to retrieve and interact with this new version. It may not be too far off, once we can find the new way to capture that xsrfToken or _rnr_se value.
I'm a bit short on time at the current moment, but would love to get this working again. It's probably a matter of messing with firebug, etc. to watch how the session gets started in browser via javascript and have PGV mimic the new URLs, etc.
Per Ward Mundy:
New version of gvoice command line sms text messaging is available, which is fixed to work with Google's new modernized "AngularJS" gvoice web interface. It was a small change to get it working, in case anyone is wondering.
Paste these commands into your shell to upgrade:
cd ~
git clone https://github.com/pettazz/pygooglevoice
cd pygooglevoice
python setup.py install
cp -p bin/gvoice /usr/bin/.
pip install --upgrade BeautifulSoup
https://pbxinaflash.com/community/threads/sms-with-google-voice-is-back-again.19717/page-2#post-129617
Related
I'm trying to send an email using SendGrid within Azure Machine Learning. This is initially just a basic test email to ensure it is working correctly.
The steps I have taken:
Pip installed SendGrid;
Zipped the SendGrid download and uploaded as a dataset to AML platform;
Attempted to run the example SendGrid Python code (which can be seen below):
I have copied steps in similar posts concerning uploading modules to AML here and here as well as ensuring the correct settings for the SendGrid API key were established on setup here.
def azureml_main():
import sendgrid
from sendgrid import SendGridAPIClient
from sendgrid.helpers.mail import Mail
message = Mail(
from_email='xxx#xyz.com',
to_emails='xxx#xyz.com',
subject='Sending with Twilio SendGrid is Fun',
html_content='<strong>and easy to do anywhere, even with Python</strong>')
try:
sg = SendGridAPIClient(os.environ.get('SG API Code'))
response = sg.send(message)
print(response.status_code)
print(response.body)
print(response.headers)
except Exception as e:
print(e)
No error message is returned in the terminal. To me this indicates there weren't any issues with the code, yet no emails have been received/sent.
python ScheduleRun.py
azureuser#will1:~/cloudfiles/code/Users/will/Schedule$
azureuser#will1:~/cloudfiles/code/Users/will/Schedule$ python ScheduleRun.py
azureuser#will1:~/cloudfiles/code/Users/will/Schedule$
Twilio SendGrid developer evangelist here.
When you run the code nothing is printed, which is concerning as neither the success or error prints are called.
If the code you have shared is the entirety of the file then I think the code is simply not running. You have defined a function but you have not called the function itself.
Try calling azureml_main() on the last line of the file:
print(response.headers)
except Exception as e:
print(e)
azureml_main()
One other thing that concerns me is this line:
os.environ.get('SG API Code')
It looks as though you intend to actually place your SG API Key in the call to os.environ.get. Instead, that string should be the identifier for an environment variable that you have set. For example, you should set the SendGrid API key as an environment variable called SENDGRID_API_KEY and then in your code fetch the API key by calling os.environ.get("SENDGRID_API_KEY"). Check out this post on working with environment variables in Python for more.
I'm using the following Python code from Slack's "Migrating to 2.x" github docs
from slackclient import SlackClient
slack_token = os.environ["SLACK_API_TOKEN"]
client = SlackClient(slack_token)
def say_hello(data):
if 'Hello' in data['text']:
channel_id = data['channel']
thread_ts = data['ts']
user = data['user']
client.api_call('chat.postMessage',
channel=channel_id,
text="Hi <#{}>!".format(user),
thread_ts=thread_ts
)
if client.rtm_connect():
while client.server.connected is True:
for data in client.rtm_read():
if "type" in data and data["type"] == "message":
say_hello(data)
else:
print "Connection Failed"
For the SLACK_API_TOKEN, I am using the Bot User OAuth Access Token for my app, found here:
The error I am getting is the following:
Failed RTM connect
Traceback (most recent call last):
File "/Users/.../slackbot/slackbot_env/lib/python3.8/site-packages/slackclient/client.py", line 140, in rtm_connect
self.server.rtm_connect(use_rtm_start=with_team_state, **kwargs)
File "/Users/.../slackbot/slackbot_env/lib/python3.8/site-packages/slackclient/server.py", line 168, in rtm_connect
raise SlackLoginError(reply=reply)
slackclient.server.SlackLoginError
Connection Failed
Why am I getting this error?!?!?!
Other context:
I am on a Mac, unlike others who have had issues online using Windows
machines.
I am running the code locally, in a virtual env, via
python script.py in my terminal.
I last successfully ran this in December, and have seen that Slack dropped support for the RTM API (?) Dec 31st 2019?
The app has been reinstalled to my workspace, and the keys did not change.
I think it may be something I need to configure/change/set/refresh on the api.slack.com/apps side, since it broke without any code changes occurring.
Why am I focusing on debugging the example for 1.x? My code was previously working using rtm_connect / 1.x using the same commands as the example code, and without any code changes it has stopped working. My code and the example code yield the same errors, so I'm using the sample code to make debugging easier. I'd like to fix this before starting the process of migrating to 2.x, so I can start with working code before embarking on a long series of changes that can introduce their own errors.
I do not think this issue is related to the Bot User OAuth Access Token, in my view you are using the right one (xoxb-). However, this issue might be related to the Slack App. Note that RTM isn't supported for the new Slack App granular scopes (see python client issue #584 and node client issue #921). If you want to use RTM, you should create rather a classic slack app with the OAuth Scope bot.
I not sure if this is the reason, but I ran into the same issues before.
The answer I found on the Slack Github is that new xoxob-* doesn't support RTM.
Please reference this web:
- https://github.com/slackapi/python-slackclient/issues/326.
So I use my OAuth Access Token instead of Bot User OAuth Access Token.
I'm using OVH API along with python wrapper:
https://pypi.python.org/pypi/ovh
When trying to execute this code:
import ovh
client = ovh.Client()
# Print nice welcome message
print "Welcome", client.get('/me')['firstname']
I get this error:
Traceback (most recent call last):
File "index.py", line 6, in <module>
print "Welcome", client.get('/me')['firstname']
File "/home/rubinhozzz/.local/lib/python2.7/site-packages/ovh/client.py", line 290, in get
return self.call('GET', _target, None, _need_auth)
File "/home/rubinhozzz/.local/lib/python2.7/site-packages/ovh/client.py", line 419, in call
raise BadParametersError(json_result.get('message'))
ovh.exceptions.BadParametersError: Invalid signature
My info is saved in the ovh.conf as the documentation suggests.
[default]
; general configuration: default endpoint
endpoint=ovh-eu
[ovh-eu]
application_key=XXXlVy5SE7dY7Gc5
application_secret=XXXdTEBKHweS5F0P0tb0lfOa8GoQPy4l
consumer_key=pscg79fXXX8ESMIXXX7dR9ckpDR7Pful
It looks that I can connect but when trying to use the services like for instance "/me", the error raises!
It is difficult to reproduce the issue because it requires an application key and it seems that it is only granted to existing customers of OVH. I couldn't even see a link to an account registration page on their site.
By looking at the code of the call() method in /ovh/client.py, it seems that their server doesn't recognise the format or the content off the signature sent by your script. According to the inline documentation the signature is generated from these parameters:
application_secret
consumer_key
METHOD
full request url
body
server current time (takes time delta into account)
Since your call is identical to the example code provided on the OVH Python package web page, the last four parameters should be valid. In that case it looks like either the application secret or the customer key (or both) in your config file are not correct.
See also the documentation on OVH site under the 'Signing requests' heading. They explain how the signature is made and what it should look like.
Perhaps try to re-create a new application API to obtain new key and secret and make sure you copy them without any additional character.
I'm maintaining a Python application using the official Dropbox API. To ask the users to let my application use their Dropbox account, I use a small script using the DropboxSession class, which is clearly the same as the one we can find on this blog post :
# Include the Dropbox SDK libraries
from dropbox import client, rest, session
# Get your app key and secret from the Dropbox developer website
APP_KEY = '******'
APP_SECRET = '******'
# ACCESS_TYPE should be 'dropbox' or 'app_folder' as configured for your app
ACCESS_TYPE = 'app_folder'
sess = session.DropboxSession(APP_KEY, APP_SECRET, ACCESS_TYPE)
request_token = sess.obtain_request_token()
url = sess.build_authorize_url(request_token)
# Make the user sign in and authorize this token
print "url:", url
print "Please visit this website and press the 'Allow' button, then hit 'Enter' here."
# Python 2/3 compatibility
try:
raw_input()
except NameError:
input()
# This will fail if the user didn't visit the above URL
access_token = sess.obtain_access_token(request_token)
#Print the token for future reference
print access_token
While it's perfectly working with Python 2.7.6, it seems to fail because of Dropbox code in Python 3.4 (the raw_input problem having been dealt with). I get this error :
Traceback (most recent call last):
File "/home/scylardor/.virtualenvs/onitu3/lib/python3.4/site-packages/dropbox/session.py", line 285, in _parse_token
key = params['oauth_token'][0]
KeyError: 'oauth_token'
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "get_access_token.py", line 12, in <module>
request_token = sess.obtain_request_token()
File "/home/scylardor/.virtualenvs/onitu3/lib/python3.4/site-packages/dropbox/session.py", line 185, in obtain_request_token
self.request_token = self._parse_token(response.read())
File "/home/scylardor/.virtualenvs/onitu3/lib/python3.4/site-packages/dropbox/session.py", line 287, in _parse_token
raise ValueError("'oauth_token' not found in OAuth request.")
ValueError: 'oauth_token' not found in OAuth request.
Long story short, after having studied the faulty code, it seems that the Dropbox code searches for a string dictionary key, despite the fact that in Python 3, those keys become bytestrings (i.e. it lookups 'oauth_token', which isn't here, instead of b'oauth_token', which is here).
However, even after having fixed the code to see if that's the only issue, no luck, I get another error further in the procedure:
Traceback (most recent call last):
File "get_access_token.py", line 25, in <module>
access_token = sess.obtain_access_token(request_token)
File "/home/scylardor/.virtualenvs/onitu3/lib/python3.4/site-packages/dropbox/session.py", line 214, in obtain_access_token
response = self.rest_client.POST(url, headers=headers, params=params, raw_response=True)
File "/home/scylardor/.virtualenvs/onitu3/lib/python3.4/site-packages/dropbox/rest.py", line 316, in POST
return cls.IMPL.POST(*n, **kw)
File "/home/scylardor/.virtualenvs/onitu3/lib/python3.4/site-packages/dropbox/rest.py", line 254, in POST
post_params=params, headers=headers, raw_response=raw_response)
File "/home/scylardor/.virtualenvs/onitu3/lib/python3.4/site-packages/dropbox/rest.py", line 227, in request
raise ErrorResponse(r, r.read())
dropbox.rest.ErrorResponse: [401] 'Unauthorized'
So the faulty functions are sess.obtain_request_token() and sess.obtain_access_token(request_token). And the Python 2.7 version works fine, but I'd like to keep Python 3 compatibility.
So, does anyone know how one's supposed to make it work in Python 3 ? Could it be deliberately broken in order to make people move on to new procedures ? I could have sworn it was working with Python 3, some time ago.
Thank you for your time if you have an idea :)
edit: It seems the Dropbox SDK just isn't fully Python 3-compatible yet. So, I guess there's nothing else to do than to wait for them to update the SDK.
Try to use version 1.6
$ pip install dropbox==1.6
Better than waiting for the SDK to be compatible, you can use (or contribute to and use) the "community" fork, dropbox-py3 (here on github).
(Those quotes are big quotes. For now it's just me coding this, and just the part I need, but everyone's welcome to help. I think it's mainly identifying the few parts that are missing a ".encode" because it's mixing bytes and strings.)
I'm trying to write a program that will (among other things) get text or source code from a predetermined website. I'm learning Python to do this, and most sources have told me to use urllib2. Just as a test, I tried this code:
import urllib2
response = urllib2.urlopen('http://www.python.org')
html = response.read()
Instead of acting in any expected way, the shell just sits there, like it's waiting for some input. There aren't even an ">>>" or "...". The only way to exit this state is with [ctrl]+c. When I do this, I get a whole bunch of error messages, like
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/m/mls/pkg/ix86-Linux-RHEL5/lib/python2.5/urllib2.py", line 124, in urlopen
return _opener.open(url, data)
File "/m/mls/pkg/ix86-Linux-RHEL5/lib/python2.5/urllib2.py", line 381, in open
response = self._open(req, data)
I'd appreciate any feedback. Is there a different tool than urllib2 to use, or can you give advice on how to fix this. I'm using a network computer at my work, and I'm not entirely sure how the shell is configured or how that might affect anything.
With 99.999% probability, it's a proxy issue. Python is incredibly bad at detecting the right http proxy to use, and when it cannot find the right one, it just hangs and eventually times out.
So first you have to find out which proxy should be used, check the options of your browser (Tools -> Internet Options -> Connections -> LAN Setup... in IE, etc). If it's using a script to autoconfigure, you'll have to fetch the script (which should be some sort of javascript) and find out where your request is supposed to go. If there is no script specified and the "automatically determine" option is ticked, you might as well just ask some IT guy at your company.
I assume you're using Python 2.x. From the Python docs on urllib :
# Use http://www.someproxy.com:3128 for http proxying
proxies = {'http': 'http://www.someproxy.com:3128'}
filehandle = urllib.urlopen(some_url, proxies=proxies)
Note that the point on ProxyHandler figuring out default values is what happens already when you use urlopen, so it's probably not going to work.
If you really want urllib2, you'll have to specify a ProxyHandler, like the example in this page. Authentication might or might not be required (usually it's not).
This isn't a good answer to "How to do this with urllib2", but let me suggest python-requests. The whole reason it exists is because the author found urllib2 to be an unwieldy mess. And he's probably right.
That is very weird, have you tried a different URL?
Otherwise there is HTTPLib, however it is more complicated. Here's your example using HTTPLib
import httplib as h
domain = h.HTTPConnection('www.python.org')
domain.connect()
domain.request('GET', '/fish.html')
response = domain.getresponse()
if response.status == h.OK:
html = response.read()
I get a 404 error almost immediately (no hanging):
>>> import urllib2
>>> response = urllib2.urlopen('http://www.python.org/fish.html')
Traceback (most recent call last):
...
urllib2.HTTPError: HTTP Error 404: Not Found
If I try and contact an address that doesn't have an HTTP server running, it hangs for quite a while until the timeout happens. You can shorten it by passing the timeout parameter to urlopen:
>>> response = urllib2.urlopen('http://cs.princeton.edu/fish.html', timeout=5)
Traceback (most recent call last):
...
urllib2.URLError: <urlopen error timed out>